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Anettt [7]
3 years ago
6

"If triangle vuw is equiangular, find k and t

Mathematics
1 answer:
balu736 [363]3 years ago
5 0

The answer is b. k=64 t=52.


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En la casa de Juan, van a pedir pizza para la comida, pero cuando la van a pedir se dan cuenta que deben pedirla por perímetro o
Degger [83]

Responder:

La pizza con un perímetro de 100 cm es más grande ¿verdad?

Explicación paso a paso:

Deja que la pizza tenga forma circular.

Sea el área de la pizza = πd² / 4 y;

Perímetro de la pizza = πd

d es el diámetro de la pizza

Si la madre dice que el que tiene un perímetro de 100 cm es más grande, para estar seguros necesitamos obtener el diámetro de la pizza. La de mayor diámetro será la pizza más grande.

P = 100cm

100 = πd

d = 100 / π

d = 100 / 3,14

d = 31,85 cm

El diámetro de la pizza mamá es de 31,85 cm.

Si el padre dice que el que tiene un área de 100 cm² es más grande, obtengamos también el diámetro para estar seguros.

A = πd² / 4

100 = πd² / 4

400 = πd²

d² = 400 / π

d² = 400 / 3,14

d² = 127,39

d = √127,39

d = 11,29 cm

Por lo tanto, el diámetro de la pizza padre es de 11,29 cm.

Dado que el diámetro de la pizza madre es mayor que el de esa, la pizza con un perímetro de 100 cm es más grande, lo que demuestra que la madre tiene razón.

6 0
3 years ago
Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o
umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that  

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

So  

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now  

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)  

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)  

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1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

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Answer:

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3y - 24 = 2x ...add 24 to both sides

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3

just because

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Answer:

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